Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using mint = map<int, int>;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const char sp = ' ';
const char cmm = ',';
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;
ll mod(ll a, ll b) { return (a % b + b) % b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
void Yes() { cout << "Yes" << endl; }
void No() { cout << "No" << endl; }
void Judge(bool b) { b ? Yes() : No(); }
void YES() { cout << "YES" << endl; }
void NO() { cout << "NO" << endl; }
void JUDGE(bool b) { b ? YES() : NO(); }
ll powMod(ll b, ll e, ll m) {
  ll r = 1;
  while (e > 0) {
    if (e & 1) r = (r % m) * (b % m) % m;
    b = (b % m) * (b % m) % m;
    e >>= 1;
  }
  return r;
}
double distance(ll x1, ll y1, ll x2, ll y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
void ppp(T n) {
  cout << n << endl;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& s, const pair<T1, T2>& p) {
  return s << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& v) {
  int len = v.size();
  s << '[';
  for (int i = 0; i < len; ++i) {
    s << v[i];
    if (i < len - 1) s << ", ";
  }
  s << ']';
  return s;
}
struct aaa {
  aaa() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(15);
  };
} fastio;
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    ppp(1);
    ppp("1 1");
    return 0;
  }
  if (n == 2) {
    ppp(2);
    ppp("1 1");
    ppp("1 2");
    return 0;
  }
  int m = (n + 2) / 2;
  ppp(m);
  for (int i = 0; i < (m); ++i) {
    cout << 1 << sp << i + 1 << endl;
  }
  for (int i = 0; i < (n - m); ++i) {
    cout << i + 2 << sp << m << endl;
  }
  return 0;
}